There are three bags `B_(1),B_(2),B_(3),B_(1)` contians 5 red and 5 green balls. `B_(2)` contains 3 red and 5 green balls and `B_3` contains 5 red and 3 green balls, bags `B_(1),B_(2)` and `B_(3)` have probabilities 3/10, 3/10, and 4/10
respectively of bieng chosen. A bag is selected at randon and a ball is randomly chosen from the bag. then which of the following options is/are correct?

Options

A. Probability that the chosen ball is green equals `(39)/(80)`
B. Probability that the chosen all is green, gen that selected bag is `B_(3)` equals `(3)/(8)`
C. Probability that the selected bag is `B_(3)`, given that the chosen ball is green equals `(4)/(13)`
D. Probability that the selected bag is `B_(3)` given that the chosen ball is green equals `(3)/(10)`

Answer

A, B, C

Explanation

`{:(,"Bag"_(1),"Bag"_2,"Bag"_(3)),("Red balls",5,3,5),("Green balls",5,5,3),("Total",10,8,8):}`
(A) P(Ball is green) `=P(B_(1))P(G//B_(1))+P_(B_(2))P(G//B_(2))+P_(B_(3))P_(G//B_(3))`
`=(3)/(10)xx(5)/(10)+(3)/(10)xx(5)/(8)+(4)/(10)xx(3)/(8)=(39)/(80)`
B(). P (ball chosen is green/ball is fro `3^(rd)` bag) `=(3)/(8)`
(C,D) P (Ball is from `3^(rd)` bag/ball chosen is green)
`=(P_(B_(3))P(G//B_(3)))/(P(B_(1))P_(G//B_(1))+P(B_(2))P(G//B_(2))+P(B_(3))P(G//B_(3)))`
`P(B_(1))=(3)/(10)`
`P(B_(2))=(3)/(10)`
`P(B_3)=(4)/(10)`
`=((4)/(10)xx(3)/(8))/((3)/(10)xx(5)/(10)+(3)/(10)xx(5)/(8)+(4)/(10)xx(3)/(8))=(4)/(13)`